1)用角的關(guān)系做簡(jiǎn)單
(a^2+c^2-b^2)/(a^2+b^2-c^2)=C/(2a-c)
【2ac(a^2+c^2-b^2)/2ac】/【2ab(a^2+b^2-c^2)/2ab】=sinC/(2sinA-sinC)
ccosB/bcosC=sinC/(2sinA-sinC)
sinCcosB/sinBcosC=sinC/(2sinA-sinC)
cosB(2sinA-sinC)=sinBcosC
2cosBsinA=cosBsinC+sinBcosC=sin(B+C)=sinA
cosB=1/2;
SinB+sin(C-A)=2Sin2A
sin(A+C)+sin(C-A)=2sin2A
2sinCcosA=4sinAcosA
sinC=2sinA
c=2a
(cosA不等于0,若等于,則sinC=2);
由cosB=1/2和c=2a不能確定一個(gè)三角形,所以無(wú)解!
2)f(0)=a+bsin0+cos0=1
a=0
f(π/4)=a+bsin(π/2)+cos(π/2)=1
a+b=1,b=1
f(x)=sin2x+cos2x=√2sin(2x+π/4)
當(dāng)X屬于[0,π/4]時(shí),x=π/8時(shí),f(x)最大為√2
矛盾!